Problem: $g(x)=-x^4+6x^2-2x-3$. On which intervals is the graph of $g$ concave down? Choose 1 answer: Choose 1 answer: (Choice A) A $-2<x<2$ only (Choice B) B $x<-1$ and $x>1$ (Choice C) C $x<-2$ and $x>2$ (Choice D) D $-1<x<1$ only
We can analyze the intervals where $g$ is concave up/down by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=-12(x+1)(x-1)$. $g''(x)=0$ for $x=-1,1$. Since $g''$ is a polynomial, it's defined for all real numbers. Therefore, our points of interest are $x=-1$ and $x=1$. Our points of interest divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $x< \llap{-}1$ $\llap{-}1<x<1$ $x>1$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $x<-1$ $x=-2$ $g''(-2)=-36<0$ $g$ is concave down $\cap$ $-1<x<1$ $x=0$ $g''(0)=12>0$ $g$ is concave up $\cup$ $x>1$ $x=2$ $g''(2)=-36<0$ $g$ is concave down $\cap$ In conclusion, the graph of $g$ is concave down over the intervals $x<-1$ and $x>1$.